Find the volume of the solid generated by revolving the region in the first quadrant bounded by the coordinate axes, the curve $y=e^x$ and the line $x=\ln 5$, rotated about the line $x= \ln 5$
Here is my work:
$$V= 2\pi \int _{a}^{b}(\text{shell radius})(\text {shell height})dx$$
$$V= 2\pi \int _{0}^{\ln 5}(x)(e^x)dx$$
$$u=x\,; du=dx\,\,\,\,\,dv=e^xdx\,;v=e^x$$
Using the integration by parts formula: $\int udv = uv-\int vdu$
$$2\pi [ xe^x|_{0}^{\ln 5} – \int_{0}^{\ln 5}e^xdx]$$
$$2\pi [ xe^x|_{0}^{\ln 5} – e^x|_{0}^{\ln 5}]$$
$$2\pi [(\ln 5)5 – (5-1)]$$
$$2\pi [5\ln 5 – 4]$$
This was the answer I got but the answer on the online homework says the correct answer is $2\pi(4-\ln5)$. Are these two answers equivalent or did I do something wrong? Any help is appreciated!
Best Answer
The radius of the shell at $x$ isn’t $x$: that’s the distance to the $y$-axis, and you’re not revolving the region about the $y$-axis. The distance from the shell at $x$ to the axis at $\ln 5$ is $\ln 5-x$. The rest of your integral is fine, so you should have
$$2\pi\int_0^{\ln 5}(\ln 5-x)e^x\,dx\;.$$
Note that all you have to do now is compute
$$2\pi\ln 5\int_0^{\ln 5}e^x\,dx$$
and combine it properly with the erroneous value that you already calculated.